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code words

by success1111 » Wed Apr 22, 2009 10:27 pm
If a code word is defined to be a sequence of different letters chosen from the 10 letters ,B,C,D,E,F,G,H,I,and J,what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A. 5 to 4
B. 3 to 2
C. 2 to 1
D 5 to 1
E 6 to 1
Last edited by success1111 on Tue Apr 28, 2009 1:28 am, edited 2 times in total.
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by scoobydooby » Wed Apr 22, 2009 10:32 pm
no of 5-letter codes: 10*9*8*7*6 (1st letter chosen in 10 ways, 2nd in 9 ways..so on as different letters must be chosen)

no of 4 letter codes: 10*9*8*7

ratio of 5 letter codes to 4 letter codes: 10*9*8*7*6 /10*9*8*7
=6/1

hence, E
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by fleshins » Wed Apr 22, 2009 10:32 pm
Isn't it 10c5/10c4? Unfortunately that doesn't match any answer ;)
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Re: code words

by Vemuri » Wed Apr 22, 2009 10:41 pm
This is a permutations question, since order matters in this case.

Number of 5 letter words that can be formed = 10p5 ==> 10*9*8*7*6.

Number of 4 letter words that can be formed = 10p4 ==> 10*9*8*7

The ratio of the number of 5-letter code wors to the number of 4-letter code words = (10*9*8*7*6)/(10*9*8*7) ==> 6:1

Answer should be E
Last edited by Vemuri on Wed Apr 22, 2009 10:49 pm, edited 2 times in total.
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by scoobydooby » Wed Apr 22, 2009 10:41 pm
@ fleshins, its more like 10P5/10P4. the order/arrangement is important here. as abcde is different from dcbae. so it is permutation here. :)
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by rossmj » Thu Apr 23, 2009 6:03 am
E.

Number of ways to pick 5 letters = 10C5 = 252
Number of ways to pick 4 letters = 10C4 = 210

However because order matter there are 5! ways to arrange the 5 letters chosen, and 4! ways to arrange the 4 letters chosen. At this point the math gets pretty long, which usually implies there is a shortcut that you are missing. If you know how to do these types of problems you can see:

10C5*5! this means that one of the 5! in the bottom will be canceled out and similarly the 4! in the 4 letters chosen will cancel out leaving us with,

10!/5!=10*9*8*7*6
and
10!/6!=10*9*8*7

leaving you with 6:1
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by gmat740 » Thu Apr 23, 2009 7:39 am
@ fleshins, its more like 10P5/10P4. the order/arrangement is important here. as abcde is different from dcbae. so it is permutation here.
Thanks a lot, I messed up with the same confusion!
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by vittalgmat » Sun Apr 26, 2009 3:20 pm
IMO the answer is 10P5/10P4, which turns out to be 6:5.
none of the answers match.

Pls check your OA ?
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by success1111 » Sun Apr 26, 2009 10:55 pm
vittalgmat wrote:IMO the answer is 10P5/10P4, which turns out to be 6:5.
none of the answers match.

Pls check your OA ?
OA is E.
Trust but verify.
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by vittalgmat » Sun Apr 26, 2009 11:19 pm
I seriously doubt the OA.

Since code words are involved, Order makes a difference. So Permutation is the name of the game and if u calculate 10P5/10P4, the ratio turns out to be 6:5.

How can u get 6:1 ??
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by Ian Stewart » Mon Apr 27, 2009 5:39 am
vittalgmat wrote:I seriously doubt the OA.

Since code words are involved, Order makes a difference. So Permutation is the name of the game and if u calculate 10P5/10P4, the ratio turns out to be 6:5.

How can u get 6:1 ??
Your method is correct, but I think you may have made a cancellation error when arriving at the ratio.

10P5 = 10*9*8*7*6
10P4 = 10*9*8*7

so when you divide 10P5 by 10P4, you get 6, and the ratio is 6 to 1.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by vittalgmat » Mon Apr 27, 2009 12:25 pm
Oh my!!! gosh...such a silly mistake.. and I kept insisting that I was right!! :-)
Sorry for all the trouble.

must have been smoking something !! ;-)
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